Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 11234, MAXM = 32;
inline void swp(int i, int j, vector<pair<int, int> > &seq) {
  seq.push_back(make_pair(i, j));
  seq.push_back(make_pair(j, i));
  seq.push_back(make_pair(i, j));
}
struct gauss {
  bitset<32> e[MAXN];
  int pos[MAXM];
  vector<pair<int, int> > seq;
  void print(int n, int m = 32) const {
    for (int i = 0; i < int(n); i++)
      for (int j = 0; j < int(m); j++)
        printf("%d%c", int(e[i][j]), " \n"[j + 1 == m]);
    puts("-----");
  }
  int reduce(int n, int m = 32) {
    int r = 0;
    for (int c = 0, i, j, k; c < m && r < n; ++c) {
      for (k = r; k < n && !e[k][c]; ++k)
        ;
      if (k >= n)
        continue;
      else if (k != r) {
        swap(e[k], e[r]);
        swp(k, r, seq);
      }
      pos[r] = c;
      for (i = 0; i < n; ++i)
        if (i != r && e[i][c]) {
          e[i] ^= e[r];
          seq.push_back(make_pair(i, r));
        }
      r++;
    }
    return r;
  }
} x, y;
int xx[11234], yy[11234];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < int(n); i++) scanf("%d", xx + i);
  for (int i = 0; i < int(n); i++) scanf("%d", yy + i);
  for (int i = 0; i < int(n); i++) {
    x.e[i] = xx[i];
    y.e[i] = yy[i];
  }
  int rx = x.reduce(n);
  int ry = y.reduce(n);
  if (rx < ry) {
    puts("-1");
    return 0;
  }
  int(&pos)[MAXM] = x.pos;
  for (int i = 0; i < int(ry); i++)
    if (y.e[i] != x.e[i]) {
      if (pos[i] < y.pos[i]) {
        x.e[i] = 0;
        x.seq.push_back(make_pair(i, i));
      }
      for (int j = i + 1; j < rx; ++j) {
        if (x.e[i][pos[j]] != y.e[i][pos[j]]) {
          x.e[i] ^= x.e[j];
          x.seq.push_back(make_pair(i, j));
        }
      }
      if (x.e[i] != y.e[i]) {
        puts("-1");
        return 0;
      }
    }
  for (int i = ry; i < rx; ++i) {
    x.e[i] = 0;
    x.seq.push_back(make_pair(i, i));
  }
  int M = x.seq.size() + y.seq.size();
  printf("%d\n", M);
  for (int i = 0; i < int(x.seq.size()); i++) {
    pair<int, int> pr = x.seq[i];
    printf("%d %d\n", pr.first + 1, pr.second + 1);
    xx[pr.first] ^= xx[pr.second];
  }
  for (int j = y.seq.size() - 1; j >= 0; --j) {
    pair<int, int> pr = y.seq[j];
    printf("%d %d\n", pr.first + 1, pr.second + 1);
    xx[pr.first] ^= xx[pr.second];
  }
  for (int i = 0; i < int(n); i++) assert(xx[i] == yy[i]);
}